A. Grandma Laura and Apples

题目连接:

http://www.codeforces.com/contest/632/problem/A

Description

Grandma Laura came to the market to sell some apples. During the day she sold all the apples she had. But grandma is old, so she forgot how many apples she had brought to the market.

She precisely remembers she had n buyers and each of them bought exactly half of the apples she had at the moment of the purchase and also she gave a half of an apple to some of them as a gift (if the number of apples at the moment of purchase was odd), until she sold all the apples she had.

So each buyer took some integral positive number of apples, but maybe he didn't pay for a half of an apple (if the number of apples at the moment of the purchase was odd).

For each buyer grandma remembers if she gave a half of an apple as a gift or not. The cost of an apple is p (the number p is even).

Print the total money grandma should have at the end of the day to check if some buyers cheated her.

Input

The first line contains two integers n and p (1 ≤ n ≤ 40, 2 ≤ p ≤ 1000) — the number of the buyers and the cost of one apple. It is guaranteed that the number p is even.

The next n lines contains the description of buyers. Each buyer is described with the string half if he simply bought half of the apples and with the string halfplus if grandma also gave him a half of an apple as a gift.

It is guaranteed that grandma has at least one apple at the start of the day and she has no apples at the end of the day.

Output

Print the only integer a — the total money grandma should have at the end of the day.

Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Sample Input

2 10

half

halfplus

Sample Output

15

Hint

题意

有n个顾客买苹果,每个苹果p元

half就是这个顾客买了一半的苹果

halfplus就是这个顾客买了一半苹果,最后还送了他半个苹果

最后恰好卖完所有苹果

问你赚了多少钱

题解:

倒着推就好了

当成模拟题做就行了

代码

#include<bits/stdc++.h>
using namespace std; long long ans = 0;
string s[45];
int main()
{
int n,p;scanf("%d%d",&n,&p);
for(int i=0;i<n;i++)
cin>>s[i];
long long now = 0;
for(int i=n-1;i>=0;i--)
{
if(s[i]=="half")
{
ans+=now*p;
now=now*2;
}
else
{
ans+=p/2+now*p;
now=now*2+1;
}
}
cout<<ans<<endl;
}

Educational Codeforces Round 9 A. Grandma Laura and Apples 水题的更多相关文章

  1. Educational Codeforces Round 9 -- A - Grandma Laura and Apples

    题意: 外祖母要卖苹果,(有很多但不知道数量),最终所有苹果都卖光了! 有n个人买苹果,如果那个人是half,他就买所有苹果的一半,如果那个人是halfplus,则他买当前苹果数量的一半,Laura还 ...

  2. Codeforces Round #367 (Div. 2) A. Beru-taxi (水题)

    Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b ...

  3. Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题

    A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...

  4. Codeforces Round #115 B. Plane of Tanks: Pro 水题

    B. Plane of Tanks: Pro Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/17 ...

  5. Codeforces Round #353 (Div. 2) A. Infinite Sequence 水题

    A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/675/problem/A Description Vasya likes e ...

  6. Educational Codeforces Round 22 E. Army Creation(分块好题)

    E. Army Creation time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...

  7. Educational Codeforces Round 59 (Rated for Div. 2) (前四题)

    A. Digits Sequence Dividing(英文速读) 练习英语速读的题,我还上来昏迷一次....只要长度大于2那么一定可以等于2那么前面大于后面就行其他no 大于2的时候分成前面1个剩下 ...

  8. Educational Codeforces Round 6 A. Professor GukiZ's Robot 水

    A. Professor GukiZ's Robot   Professor GukiZ makes a new robot. The robot are in the point with coor ...

  9. Codeforces Round #327 (Div. 2) A. Wizards' Duel 水题

    A. Wizards' Duel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/prob ...

随机推荐

  1. perl6中的hash定义(2)

    use v6; , :b, :!c; say %ha; say %ha<a>; #这里不能用%ha{a}, {a}表示调用a()函数了, 在perl6中, {}有特别函义 say %ha{ ...

  2. linux网络编程之IO函数

    Linux操作系统中的IO函数主要有read(),write(),recv(),send(),recvmsg(),sendmsg(),readv(),writev(). 接收数据的recv()函数 # ...

  3. Django2.0如何配置urls文件

    刚开始学django,创建的第一个工程无法启动,后来发现是由于教程是针对较低版本的Django,我用的是Django2.0和Python3.6,两个都是发文为止的最新版本,urls文件设置方法和旧版本 ...

  4. 设计模式之笔记--组合模式(Composite)

    组合模式(Composite) 定义 组合模式(Composite),将对象组合成树形结构以表示“部分-整体”的层次结构.组合模式使得用户对单个对象和组合对象的使用具有一致性.       组合模式有 ...

  5. C#ActiveX安装项目

    C#开发的ActiveX控件发布方式有三种: 制作客户端安装包,分发给客户机安装: 制作在线安装包,客户机联机安装: 使用html中object的codebase指向安装包地址. 以下为制作安装包: ...

  6. overflow属性在IE6下面失去效果

    自然状态下 overflow的属性设置,本来是超过了一定的长度时会自动产生滚动条,但是在ie6下面失效了. 例如:原来的代码: .code{overflow-x:auto;margin:5px aut ...

  7. 使用 Visual Studio 部署 .NET Core 应用 ——.Net Core 部署到Ubuntu 16.04

    .Net Core 部署到Ubuntu 16.04 中的步骤 1.安装工具 1.apache 2..Net Core(dotnet-sdk-2.0) 3.Supervisor(进程管理工具,目的是服务 ...

  8. hdu 2686&&hdu 3376(拆点+构图+最小费用最大流)

    Matrix Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Subm ...

  9. AC日记——「SCOI2016」幸运数字 LiBreOJ 2013

    「SCOI2016」幸运数字 思路: 线性基: 代码: #include <bits/stdc++.h> using namespace std; #define maxn 20005 # ...

  10. Linux下cp命令的使用说明

    [root@www ~]# cp [-adfilprsu] 来源档(source) 目标档(destination)[root@www ~]# cp [options] source1 source2 ...